Minimum size of an orbiting object to be more than a point of light to people on the ground?

Factual Questions
21

So continuing my journey into atrocious geometry, let me dig myself even further into a hole.

Given 2 sides of 35000 feet and a base of 250 feet, I think you can just use ratios to determine the size of the base at any length of the sides. So, for instance, if the sides were 70,000 feet, the base would be 500 feet. (oh, boy, I HOPE this is right. Otherwise, I’m going to look really dumb – nothing new, now that I think of it.)

So to get an object about the size of a 747 at 35,000 feet, at, say, ISS height, we’d…

Let’s see. Using a figure of 270 miles for ISS orbital height, 270 miles is 1,425,600 feet. Divide that by 35000 and we get about 40. 250 feet times 40 is 10,000 feet.

So to get an object the apparent size of a 747 at 35,000 at an orbit of 270 miles, the object would have to be 10,000 feet big, or about 2 miles wide.

So either this is a back-of-envelope calculation for the size you need, or I just dug myself a very big hole and jumped into it. :smiley:

J.

22

Somewhere in the planning stages, people would start screaming loudly about your putting a civilization-ending impactor on a near-collision course with the Earth.

23

Using this triangle calculator,
http://www.cleavebooks.co.uk/scol/calrtri.htm
I make the length of a 747-equivalent object at ISS-height to be 3300 metres, or 2.05 miles; so you are correct.

24

A 3km object in the sky would be very bright, much brighter than the ISS, which is only 100 metres long, but is still brighter than Venus. That seems a bit of overkill if you are only trying to catch attention. How about two 100 metre objects 3 kilometers apart? They should be easily resolved into two objects by anyone with normal eyesight.

25

Fine, we’ll go grab an asteroid and tow it into LEO. What’s the worst that could happen?

26

If it helps, one way to detect the extremely weak gravitational effects of the moon is to build a gigantic detector 12,000 kilometers in diameter. With that large a detector you could detect the signal from the moon by movement of water that could be a meter or several meters high! Several meters high in a detector 12,742,000 meters across. That’s actually a pretty weak signal.

The reason you can’t detect the rock on your desk with the planet-sized detector is that the rock only has that apparent size of the moon when it’s really close. Go a hundred meters away and the rock is only a pinpoint. However the moon has nearly the same apparent diameter from anywhere on Earth, even if you were on the far side of the Earth and could look through a giant window the moon would only be a bit smaller, it would be hard to tell with your naked eye. Anyway, if you built a detector 12 meters across to detect the gravitational field of that rock you your desk you might detect a variation of 10^-6 meters, if I got my orders of magnitude correct. Of course there are so many other variables that would swamp that signal–vibrations, changes in temperature, air currents, trucks passing by outside the lab, footsteps, electromagnetic fields, and on and on and on. You’d spend a lot of time systematically identifying and eliminating sources of experimental error, but if you were careful and well funded and spent several years working on the problem…

27

For what it’s worth, there are actual scientific experiments that do measure the gravitational fields of the equivalent of paperweights (though in practice, they’re usually spheres of lead or steel, or some other precisely-chosen shape). Having seen what those experimenters have to do to pick out their signals, I can confidently state that I would go insane if that were my field.